A119980 Order of the following permutation on 3n+1 symbols. Write the 3n+1 symbols horizontally into a 3-column grid and read them off vertically, i.e., column after column.

1, 3, 6, 6, 11, 15, 52, 38, 51, 9, 360, 260, 35, 39, 364, 1932, 680, 532, 1122, 260, 2415, 3570, 168, 360, 71, 12285, 836, 12, 1680, 1155, 858, 936, 7956, 48300, 171120, 234, 4428, 235752, 712, 990, 119, 364182, 406, 11220, 412920, 25584, 476, 19998, 6486

Offset

0,2

Links

Andrew Howroyd, Table of n, a(n) for n = 0..1000

Example

For n=2, the grid with 0..6 by rows is
   0 1 2
   3 4 5      first column is one longer
   6
Reading them by columns gives (0,3,6,1,4,2,5) which as a permutation has order 6, so a(2) = 6.

Prog

(GAP) # GAP / KANT / KASH
# SpartaEncrypt(d, L) returns the list M obtained by writing L in d columns
# and then concatenating these columns
SpartaEncrypt := function(d, L)
local len, i, M;
len := Length(L);
M := [];
for i in [1..d] do
Append(M, L{[i, d+i..d*IntQuo(len-i, d)+i]});
od;
return M;
end;
# SpartaOrd(d, m) computes the order of SpartEncrypt(d, [0..m-1])
# in the group S_m
SpartaOrd := function(d, m)
local L, M, i;
M := [0..m-1];
L := [0..m-1];
i := 0;
repeat
i := i + 1;
L := SpartaEncrypt(d, L);
until L=M;
return i;
end;
d:=3; r:=1;
a := List([0..60], n->SpartaOrd(d, d*n+r));
(PARI)
P(n, w, j)={my(c=j%w); if(c==0, j/w, j\w + c*n + 1)}
Follow(s, f)={my(t=f(s), k=1); while(t>s, k++; t=f(t)); if(s==t, k, 0)}
CyclePoly(n, w, x)={my(q=0); for(i=0, w*n, my(l=Follow(i, j->P(n, w, j))); if(l, q+=x^l)); q}
a(n)={my(q=CyclePoly(n, 3, x), m=1); for(i=1, poldegree(q), if(polcoef(q, i), m=lcm(m, i))); m} \\ Andrew Howroyd, Jan 04 2024

Crossrefs

The case for 2 columns is A002326.

Cf. A003572.

Sequence in context: A113497 A210692 A158662 * A066779 A132384 A378120

Adjacent sequences: A119977 A119978 A119979 * A119981 A119982 A119983

Keyword

nonn

Author

Roland Miyamoto, Aug 03 2006

Status

approved